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If $50 \%$ of $1 \mathrm{M} \mathrm{Na}_2 \mathrm{SO}_4$ is dissociated in aqueous solution of density $1.2 \mathrm{~g} \mathrm{~mL}^{-1}$, what is the molality of $\mathrm{Na}^{+}$ion in the solution?
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The correct answer is:
0.95
Density (d) of solution $=1.2 \mathrm{~g} / \mathrm{ml}$
$$
\underset{1 \mathrm{~mol}}{\mathrm{Na}_2 \mathrm{SO}_4} \longrightarrow \underset{2 \mathrm{~mol}}{2 \mathrm{Na}^{+}}+\mathrm{SO}_4^{2-}
$$
MW of $\mathrm{Na}_2 \mathrm{SO}_4=142 \mathrm{~g} / \mathrm{mol}$
$\therefore 50 \%$ dissociation of $\mathrm{Na}_2 \mathrm{SO}_4$ produces $1 \mathrm{~mol}$ of $\mathrm{Na}^{+}$
$\therefore \quad 1(\mathrm{M}) \mathrm{Na}_2 \mathrm{SO}_4$ solution has $1(\mathrm{M}) \mathrm{Na}^{+}$ions in $1000 \mathrm{ml}$ solution $=1200 \mathrm{~g}$ solution.
$\therefore \quad$ Weight of solvent $=(1200-142)=1058 \mathrm{~g}$
$\therefore \quad$ The molality of $\mathrm{Na}^{+}$ion $=\frac{1 \times 1000}{1058}=0.95 \mathrm{~m}$
$$
\underset{1 \mathrm{~mol}}{\mathrm{Na}_2 \mathrm{SO}_4} \longrightarrow \underset{2 \mathrm{~mol}}{2 \mathrm{Na}^{+}}+\mathrm{SO}_4^{2-}
$$
MW of $\mathrm{Na}_2 \mathrm{SO}_4=142 \mathrm{~g} / \mathrm{mol}$
$\therefore 50 \%$ dissociation of $\mathrm{Na}_2 \mathrm{SO}_4$ produces $1 \mathrm{~mol}$ of $\mathrm{Na}^{+}$
$\therefore \quad 1(\mathrm{M}) \mathrm{Na}_2 \mathrm{SO}_4$ solution has $1(\mathrm{M}) \mathrm{Na}^{+}$ions in $1000 \mathrm{ml}$ solution $=1200 \mathrm{~g}$ solution.
$\therefore \quad$ Weight of solvent $=(1200-142)=1058 \mathrm{~g}$
$\therefore \quad$ The molality of $\mathrm{Na}^{+}$ion $=\frac{1 \times 1000}{1058}=0.95 \mathrm{~m}$
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